Are the cells in this image prokaryotic or eukaryotic

Ahat [919]

The greatest amount of AB would be produced if the equilibrium constant of the reaction is equal to $1.0 \;X \;10^5$ . Hence, option D is correct.

<h3>What is an equilibrium constant?</h3>

A number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

The equilibrium constant expression is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants.

If the value of K is greater than 1, the products in the reaction are favoured. If the value of K is less than 1, the reactants in the reaction are favoured.

Hence, option D is correct.

Learn more about the equilibrium constant here:

brainly.com/question/10038290

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7 0

1 year ago

The aldol condensation involves nucleophilic attack of an enolate to a carbonyl. Use the pop-up menus to identify the total numb

Sindrei [870]

Aldol condensation involves the reaction of an acid or base with a carbonyl group producing a nucleophile that attacks another carbonyl compound to yield a β-hydroxyaldehyde or β-hydroxyketone compound.

<h3>What is aldol condensation?</h3>

The aldol condensation is a reaction in organic chemistry in which there is a reaction between an acid or base and a carbonyl group which then serves as the nucleophile that attcks a second carbonyl to yield a β-hydroxyaldehyde or β-hydroxyketone compound.

The aldol condensation may be acid catalysed or base catlysed. The question is incomplete hence the complete mechanimsms can not be decuced.

Learn more about aldol condensation: brainly.com/question/9415260

3 0

2 years ago

What is the silver ion concentration in a solution prepared by mixing 369 mL 0.373 M silver nitrate with 411 mL 0.401 M sodium c

LuckyWell [14K]

Answer:

[A g + ] = 3.12 *10^-6 M

Explanation:

Step 1: Data given

Volume silver nitrate = 369 mL

Molarity silver nitrate = 0.373 M

Volume sodium chromate = 411 mL

Molarity sodium chromate = 0.401 M

The Ksp of silver chromate is 1.2 * 10^− 12

Step 2: The balanced equation

2 A g + ( a q ) + C rO4^2- ( a q ) →Ag2CrO4 ( s)

Step 3:

[Ag+]i = [AgNO3] * V1/(V1+V2) * 1 mol Ag+ / 1 mol AgNO3

[Ag+]i = 0.373 M * 0.369/ (0.369+0.411)

[Ag+]i = 0.176 M

[C rO4^2]i = [Ag2CrO4] * V2 /(V1+V2) * 1mol C rO4^2 / 1 mol Ag2CrO4

[C rO4^2i = 0.401 M * 0.411 / (0.369+0.411)

[C rO4^2]i = 0.211 M

Ksp = [Ag+]²[CrO4^2-] = 1.2 * 10^− 12

[CrO4^2-]f = [CrO4^2-]i - 0.5 * [Ag+]i

[CrO4^2-]f = 0.211 -0.088 = 0.123 M

1.2 * 10^− 12 = ( 2 x ) ²*( 0.123M + x )

[ C O 3^ −2] f >> x

1.2 * 10^− 12 = ( 2 x ) ²* 0.123M

x = 1.56 * 10^-6

[A g + ] f = 2x = 3.12 *10^-6 M

,

4 0

3 years ago

What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa

almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$